If $x \star y = 5x-y$ and $x \veebar y = x^{2}+4y^{2}$, find $5 \star (-1 \veebar -3)$.
Answer: First, find $-1 \veebar -3$ $ -1 \veebar -3 = (-1)^{2}+4(-3)^{2}$ $ \hphantom{-1 \veebar -3} = 37$ Now, find $5 \star 37$ $ 5 \star 37 = (5)(5)-37$ $ \hphantom{5 \star 37} = -12$.